YES

We show the termination of the TRS R:

  min(X,|0|()) -> X
  min(s(X),s(Y)) -> min(X,Y)
  quot(|0|(),s(Y)) -> |0|()
  quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
  log(s(|0|())) -> |0|()
  log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: min#(s(X),s(Y)) -> min#(X,Y)
p2: quot#(s(X),s(Y)) -> quot#(min(X,Y),s(Y))
p3: quot#(s(X),s(Y)) -> min#(X,Y)
p4: log#(s(s(X))) -> log#(s(quot(X,s(s(|0|())))))
p5: log#(s(s(X))) -> quot#(X,s(s(|0|())))

and R consists of:

r1: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

The estimated dependency graph contains the following SCCs:

  {p4}
  {p2}
  {p1}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: log#(s(s(X))) -> log#(s(quot(X,s(s(|0|())))))

and R consists of:

r1: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

The set of usable rules consists of

  r1, r2, r3, r4

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          log#_A(x1) = ((1,1),(0,0)) x1 + (1,1)
          s_A(x1) = ((1,1),(0,1)) x1 + (2,2)
          quot_A(x1,x2) = x1 + (1,0)
          |0|_A() = (0,1)
          min_A(x1,x2) = x1
  
    precedence:
    
      log# = s = quot = |0| > min
    
    partial status:
  
      pi(log#) = []
      pi(s) = []
      pi(quot) = [1]
      pi(|0|) = []
      pi(min) = [1]

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: quot#(s(X),s(Y)) -> quot#(min(X,Y),s(Y))

and R consists of:

r1: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

The set of usable rules consists of

  r1, r2

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          quot#_A(x1,x2) = x1 + (1,0)
          s_A(x1) = ((1,1),(0,1)) x1 + (3,1)
          min_A(x1,x2) = ((1,1),(0,1)) x1 + (1,1)
          |0|_A() = (1,1)
  
    precedence:
    
      s = min > quot# = |0|
    
    partial status:
  
      pi(quot#) = [1]
      pi(s) = []
      pi(min) = []
      pi(|0|) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.

-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: min#(s(X),s(Y)) -> min#(X,Y)

and R consists of:

r1: min(X,|0|()) -> X
r2: min(s(X),s(Y)) -> min(X,Y)
r3: quot(|0|(),s(Y)) -> |0|()
r4: quot(s(X),s(Y)) -> s(quot(min(X,Y),s(Y)))
r5: log(s(|0|())) -> |0|()
r6: log(s(s(X))) -> s(log(s(quot(X,s(s(|0|()))))))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  weighted path order
  
    base order:
  
      matrix interpretations:
      
        carrier: N^2
        order: standard order
        interpretations:
          min#_A(x1,x2) = ((0,1),(0,0)) x1 + ((0,1),(0,0)) x2 + (2,2)
          s_A(x1) = ((0,0),(0,1)) x1 + (1,1)
  
    precedence:
    
      min# = s
    
    partial status:
  
      pi(min#) = []
      pi(s) = []

The next rules are strictly ordered:

  p1

We remove them from the problem.  Then no dependency pair remains.